2x^2+x-48=0 eching

$$2 { x }^{ 2 } +x-48=0$$

x uchun yechish

$$x = \frac{\sqrt{385} - 1}{4} \approx 4,655354218$$

$$x=\frac{-\sqrt{385}-1}{4}\approx -5,155354218$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$2x^{2}+x-48=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, 1 ni b va -48 ni c bilan almashtiring.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-48\right)}}{2\times 2}$$

1 kvadratini chiqarish.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-48\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1-8\left(-48\right)}}{2\times 2}$$

-8 ni -48 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{1+384}}{2\times 2}$$

1 ni 384 ga qo'shish.

$$x=\frac{-1±\sqrt{385}}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$x=\frac{-1±\sqrt{385}}{4}$$

x=\frac{-1±\sqrt{385}}{4} tenglamasini yeching, bunda ± musbat. -1 ni \sqrt{385} ga qo'shish.

$$x=\frac{\sqrt{385}-1}{4}$$

x=\frac{-1±\sqrt{385}}{4} tenglamasini yeching, bunda ± manfiy. -1 dan \sqrt{385} ni ayirish.

$$x=\frac{-\sqrt{385}-1}{4}$$

Tenglama yechildi.

$$x=\frac{\sqrt{385}-1}{4} x=\frac{-\sqrt{385}-1}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$2x^{2}+x-48=0$$

48 ni tenglamaning ikkala tarafiga qo'shish.

$$2x^{2}+x-48-\left(-48\right)=-\left(-48\right)$$

O‘zidan -48 ayirilsa 0 qoladi.

$$2x^{2}+x=-\left(-48\right)$$

0 dan -48 ni ayirish.

$$2x^{2}+x=48$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2x^{2}+x}{2}=\frac{48}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{1}{2}x=\frac{48}{2}$$

48 ni 2 ga bo'lish.

$$x^{2}+\frac{1}{2}x=24$$

\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{4} olish uchun. Keyin, \frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=24+\left(\frac{1}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{4} kvadratini chiqarish.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=24+\frac{1}{16}$$

24 ni \frac{1}{16} ga qo'shish.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{385}{16}$$

x^{2}+\frac{1}{2}x+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{385}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{385}{16}}$$

Qisqartirish.

$$x+\frac{1}{4}=\frac{\sqrt{385}}{4} x+\frac{1}{4}=-\frac{\sqrt{385}}{4}$$

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

$$x=\frac{\sqrt{385}-1}{4} x=\frac{-\sqrt{385}-1}{4}$$